Comparing between two arrays, order matters-CodePudding

week one with Python. I would like to compare the value of two arrays where order matters and print out text. Here's what I have:

a =[2, 3, 4, 2, 6]
b =[1, 2, 3, 4, 5]
c =[0, 1, 2, 3, 4]

compare = map(lambda x, y: x > y, a, b)
print(list(compare))

and I get:

[True, True, True, False, True]

Rather than printing the list(compare) True and False I would like to print 'list a is below list b' if False. It only needs to print once if more than one False is present. I've tried many print statements without success.

if compare == False:
    print ('list a is below list b')

nothing

if compare is False:
    print ('list a is below list b')

crickets

I've tried many other ways.

Is there also a way to do a loop comparing if a>b, a>c and b>c then print('list b is below list c'). Order between lists matters. I could do the compare = map lambda 3 times but I was wondering if there's a way to loop it?

Thank you!

CodePudding user response：

If you want to check compare has at least one value False, just use the in operator.

if False in compare:
    do action

CodePudding user response：

You can compare 3 lists at the same time and use all(). all() function returns True iff all conditions in the iterator provided to it are True. -

a =[2, 3, 4, 2, 6]
b =[1, 2, 3, 4, 5]
c =[0, 1, 2, 3, 4]

compare = map(lambda x, y, z: x > y and y > z, a, b, c)
print(not any(compare))

outputs -

False

because the 3rd index is False.

If I update the a -

>>> a =[2, 3, 4, 5, 6]
>>> b =[1, 2, 3, 4, 5]
>>> c =[0, 1, 2, 3, 4]
>>> compare = map(lambda x, y, z: x > y and y > z, a, b, c)
>>> print(all(compare))
True